Seattle daily post-intelligencer. [volume] (Seattle, W.T. [Wash.]) 1881-1888, 1887-05-15
WEIGHT or CATTLE by MEA-
SURENENT.

An English journal. IN answer to a
correspondent, gives the manner Of
arriving AZ the weight Of cattle BIZ me2-
Suremert- To find the weight OF 4 fat
animal the square Of the girth Is null
tlplled By length. and the product BY
-. The explanation. which viii be
interesting 70 many feeders and breed
ers. IS as follows
The length and depth OF the animal
oe1ng nleas1red-the first from the
shonieer-tOp 70 the taii-head. ard the
second Immediately behind the shout
csr-these dimensions bring the figure
Of the animal into the form Of 3 cylin-
Ger, or nearly 50. The rule for finding
the contents Of q cylinder Is TO find the
area Of the end and 70 multiply that
sum BY the length. The common
method Is 70 multiply the square OF
tue diameter by ISL, 1.-THE area Of
circle whose diameter IS unitv) and
thls product by the length for the solid
contents.
But in measuring cattle the girth or
circumference. and not the diameter.
i obtained: and as the rule for finding
the diameter correctly from the cir
conference involves itself into long
decimal mulfipiiers, the process, eSpe-
pecially when feet and odd inches are
the dimensions, Is complicated aud te-
diOus- The more simple method is,
therefore, to multiply the square of the
circumference BY 0.795T75,s and that
product multiplied by the length gives
the contents; which again multiplied
BY the established weight OF q cubic foot
or other measure VIII give the weight
OF the animal
To find the proportional weight Of a
cubic foot: Find BY the above rule the
number Of cubic feet which the animal
contains. and weigh the four quarters
after il IS killed; and the former divid-
ed by the latter will give the weight
per cubic foot. Thus. il an OK mea-
sures feet girth, 6 feet length: plus
$ equals "a plus .0795775 equals :. 09298
plus equals 30. 55765 cubic feet in the
animal;s and if the forequarters OF the
killed animal weighed IS Smithfeld
stones. 81 pounds {735ll pounds, at s
pounds per stonel. This weight, di-
vided by the number of cubic feet in
the animal. gives the weight of cubic
foot. Hence. 81 stones 81 pounds
equal 9l.4642S57; divided by 1856.
give 1893 stones per cubic foot. And
this is the actual weight assumed for a
cubic foot.
The calculation may he shortened.
As .0795775 and 1888 are both constant
multipliers in the operation. they may
be multiplied together, and the product
used in one muitiplier. thus: 0795775
multiplied BY 1888 gives .23sl754675.
But .23s. or three figures, may be near
enough for multiplier. Thus times
is 81, times 6 is 32I, times .2Ss gives
81 .392, or 81 stones, 81 pounds. In
place of .23S some use "a, which gives
a higher weight. Thus. then, to find
the weight of a fat animal multiply
the square of the girth by the length,
and that product by obs; or take 22s-
l000th parts of il, or use any lower or
more convenient denomination of the
same value.
Another popular rule is to multiply
the square of the girth by five times
the length, and divide the product by
p. TO get the weight of the four quar-
ters-i. e., multiply the square Of the
girth BY the length, and take the 5-2lst
part of the product for the weight
Now 5-2l converted into deCimals is
.23S09523. which exactly agrees. in so
far rs the decimal numbers necessary
fer the calculation are required, with
the numbers given in the other rule.
%This number Is obtained in the fol-
lowing way: The area OF circle EQUALS
the square of the circumference diyid-
ed by four times 3.l4l6, tthe circum-
ference of a circle whose diameter IS
unitvl or divided i25564 (i4x3 l4l6)
0795775. Hence the square of the cir
conference multiplied by .0795775 gives
the area of the circle.